Biased Coin

A biased coin with the probability of getting a head is 70% of the time and tail 30%. That is, P(H) = 0.7 and P(T) = 0.3. Toss the coin THREE times. Find the following probabilities: P(HHT) and P(2 Heads).

First notice that the two probabilities are not the same. P(HHT) asks for the probability of getting a sequence of HHT, while P(2 Heads) asks for the probability of getting exactly 2 heads. There are 3 ways to get exactly 2 heads in 3 tosses: HHT, HTH, and THH. Also, note that the two events (whether we get a H or a T) are independent.

P(HHT) = P(H)P(H)P(T) = (.7)(.7)(.3) = .147.
P(2Hs) = P(HHT) + P(HTH) + P(THH) = .147 + .147 + .147 = .441

Normal Distribution

This simulation of Normal Distribution is shown at Cité des Sciences et de l'Industrie.

http://www.cite-sciences.fr/fr/cite-des-sciences/

Implicit Differentiation

Find the value of $\frac{d^2y}{dx^2}$ of $x^3 + y^3 = 56$ at $(-2,4)$.

Start by differentiating implicitly both sides of the equation with respect to x.  $3x^2 + 3y^2 y' = 0$. Solve for y': $y' = \dfrac{-3x^2}{3y^2} = \dfrac{-x^2}{y^2}$. But this is the first derivative. To find the second derivative, differentiate (implicitly) one more time, now using the quotient rule:
$y'' = \dfrac{(-2x) (y^2) - (-x^2) 2y y'}{(y^2)^2}$. Plug in y':
$y'' = \dfrac{-2xy^2 + 2x^2 y (-x^2/y^2)}{y^4}$.
At this point, we plug in (-2,4). No need to simplify:
$\dfrac{d^2y}{dx^2} |_{(-2,4)} = \dfrac{-2(-2)(4)^2 + 2(-2)^2 (4) (-(-2)^2/4^2)}{4^4}= \dfrac{7}{32}$.

Slope of a line

Given 2 points $(x_1, y_1)$ and $(x_2, y_2)$, the slope of a line that goes through these two points is a ratio of change in y to change in x. That is,
$\dfrac{y_2 - y_1}{x_2 - x_1}$
http://www.learningwave.com/lwonline/algebra_section2/slope3.html

Fibonacci Sequence

1, 1, 2, 3, 5, 8, 13, ...
$a_n = a_{n-1} + a_{n-2}$